What is the sum from n=1 to infinity of 8/(n^2-1) ?

The sum from n=1 to infinity of 8/(n^2-1) is 6

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sum from n=1 to infinity of 8/(n^2-1)

n = 1 \sum \infty \frac{8}{n ^{2} - 1}

6

Solution steps

n = 2 \sum \infty \frac{8}{n ^{2} - 1}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 8 \cdot n = 2 \sum \infty \frac{1}{n ^{2} - 1}

Take the partial fraction of

\frac{1}{n ^{2} - 1} : - \frac{1}{2 ( n + 1 )} + \frac{1}{2 ( n - 1 )}

= 8 \cdot n = 2 \sum \infty - \frac{1}{2 ( n + 1 )} + \frac{1}{2 ( n - 1 )}

Expand telescoping series:

\frac{3}{4}

Simplify

8 \cdot \frac{3}{4} : 6

= 6

k = 1 \sum \infty \frac{3}{4 ^{k}}

n = 3 \sum \infty \frac{1}{n ^{2} + 5 n}

n = 1 \sum \infty \frac{1}{n ^{3} + 4}

n = 0 \sum \infty n^{- \frac{2}{3}}

n = 1 \sum \infty 3 (- \frac{1}{8})^{3 n}

What is the sum from n=1 to infinity of 8/(n^2-1) ?
The sum from n=1 to infinity of 8/(n^2-1) is 6